Prove

Let $(x+3)=A \frac{d}{d x}\left(x^{2}-2 x-5\right)+B$

$(x+3)=A(2 x-2)+B$

Equating the coefficients of x and constant term on both sides, we obtain

$2 A=1 \Rightarrow A=\frac{1}{2}$

$-2 A+B=3 \Rightarrow B=4$

$\therefore(x+3)=\frac{1}{2}(2 x-2)+4$

$\Rightarrow \int \frac{x+3}{x^{2}-2 x-5} d x=\int \frac{\frac{1}{2}(2 x-2)+4}{x^{2}-2 x-5} d x$

$=\frac{1}{2} \int \frac{2 x-2}{x^{2}-2 x-5} d x+4 \int \frac{1}{x^{2}-2 x-5} d x$

Let $I_{1}=\int \frac{2 x-2}{x^{2}-2 x-5} d x$ and $I_{2}=\int \frac{1}{x^{2}-2 x-5} d x$

$\therefore \int \frac{x+3}{\left(x^{2}-2 x-5\right)} d x=\frac{1}{2} I_{1}+4 I_{2}$   ...(1)

Then, $I_{1}=\int \frac{2 x-2}{x^{2}-2 x-5} d x$

Let $x^{2}-2 x-5=t$

$\Rightarrow(2 x-2) d x=d t$

$\Rightarrow I_{1}=\int \frac{d t}{t}=\log |t|=\log \left|x^{2}-2 x-5\right|$     ...(2)

$\begin{aligned} I_{2} &=\int \frac{1}{x^{2}-2 x-5} d x \\ &=\int \frac{1}{\left(x^{2}-2 x+1\right)-6} d x \\ &=\int \frac{1}{(x-1)^{2}+(\sqrt{6})^{2}} d x \\ &=\frac{1}{2 \sqrt{6}} \log \left(\frac{x-1-\sqrt{6}}{x-1+\sqrt{6}}\right) \end{aligned}$     ...(3)

Substituting (2) and (3) in (1), we obtain

$\int \frac{x+3}{x^{2}-2 x-5} d x=\frac{1}{2} \log \left|x^{2}-2 x-5\right|+\frac{4}{2 \sqrt{6}} \log \left|\frac{x-1-\sqrt{6}}{x-1+\sqrt{6}}\right|+C$

$=\frac{1}{2} \log \left|x^{2}-2 x-5\right|+\frac{2}{\sqrt{6}} \log \left|\frac{x-1-\sqrt{6}}{x-1+\sqrt{6}}\right|+\mathrm{C}$